Answer:
298.9m
Explanation:
Given that : Note that the object exploded into two.
Let mass of lighter fragment be m
Let mass of heavier fragment = 7m
Let the velocity of lighter fragment = V1
Let the velocity of heavier fragment = V2
Recall that object was at rest before explosion occur. Hence, momentum of system is zero.
mv1 + 7mv2 = 0
V1 = -7V2
force of Friction on lighter fragment
f1 = μmg
force of friction on heavier fragment
f2 = 7μmg
Deceleration of lighter fragment = a1
we know that force of friction is given as
ma1 = - f1
ma1 = -μmg
a1 = -μmg/m
a1= -μg
Deceleration of heavier fragment = a2
7ma2 = -f2
7ma2 = - μ7mg
a2 = -μg
Let the final velocity of lighter fragment = V3 = 0 m/s
Let the final velocity of heavier fragment = V4 = 0 m/s
Distance traveled by lighter fragment before rest position is = d1
Distance traveled by the heavier fragment before rest position is = d2 = 6.10 m
From :
V^2 = u^2 + 2as
V4^2 = v2^2+2a2d2
d2 = v2^2/2μg ............ (1)
V3^2 = v1^2+2a1d1
0 = (-7v2)^2+2(-μg)d1
0 = 49v2^2+2(-μg)d1
Make d1 the subject
0 = 49v2^2-2(μg)d1
2(μg)d1 = 49v2^2
d1 = 49v2^2/2μg............ (2)
Put equation (1) into (2)
d1 = 49d2
d1 = 49×6.1
d1 = 298.9m
